In the curve `y=x^(3)+ax and y=bx^(2)+c` pass through the point `(-1,0)` and have a common tangent line at this point then the value of `a+b+c` is

Options

A. 0
B. 1
C. `-3`
D. `-1`

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the necessary values for \( a \), \( b \), and \( c \).

### Step 1: Substitute the point into the first equation
The first curve is given by:
\[ y = x^3 + ax \]
We know that it passes through the point \((-1, 0)\). Substituting \( x = -1 \) and \( y = 0 \):
\[ 0 = (-1)^3 + a(-1) \]
This simplifies to:
\[ 0 = -1 - a \]
Rearranging gives:
\[ a = -1 \]

### Step 2: Substitute the point into the second equation
The second curve is given by:
\[ y = bx^2 + c \]
Again, substituting \( x = -1 \) and \( y = 0 \):
\[ 0 = b(-1)^2 + c \]
This simplifies to:
\[ 0 = b + c \]
Thus, we have:
\[ b + c = 0 \quad \text{(1)} \]

### Step 3: Find the derivatives of both equations
Next, we need to find the derivatives of both curves to ensure they have a common tangent at the point \((-1, 0)\).

For the first curve:
\[ \frac{dy}{dx} = 3x^2 + a \]
Substituting \( a = -1 \) and \( x = -1 \):
\[ \frac{dy}{dx} = 3(-1)^2 - 1 = 3 - 1 = 2 \]

For the second curve:
\[ \frac{dy}{dx} = 2bx \]
Substituting \( x = -1 \):
\[ \frac{dy}{dx} = 2b(-1) = -2b \]
Setting the derivatives equal for the common tangent:
\[ 2 = -2b \]
Solving for \( b \):
\[ b = -1 \]

### Step 4: Substitute \( b \) back into equation (1)
Now we substitute \( b = -1 \) into equation (1):
\[ -1 + c = 0 \]
This gives:
\[ c = 1 \]

### Step 5: Calculate \( a + b + c \)
Now we have:
- \( a = -1 \)
- \( b = -1 \)
- \( c = 1 \)

Calculating \( a + b + c \):
\[ a + b + c = -1 - 1 + 1 = -1 \]

### Final Answer
The value of \( a + b + c \) is:
\[ \boxed{-1} \]

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